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There are three players with three alternatives A,B, and C. Players simultaneously vote and majority wins. If no majority then A wins. Payoffs are:

$U_1(A) = U_2(B) = U_3(C) = 2$
$U_1(B) = U_2(C) = U_3(A) = 1$
$U_1(C) = U_2(A) = U_3(B) = 0$

The example states (A,B,C), (A,A,A), and (A,B,A) are all NE. I get why these are NE (I can reason this out) but I do not understand the process to get here. I could write out all possible strategy profiles and try to reason each one out but there has to be a better and efficient way. I also cannot draw the matrix for this game as I usually would to solve for NE to find best response. The preferences here also violate transitivity, which is just adding to my confusion.

My question is how would I go about finding all NE?

Are there more NE than these 3 the example mentions?

How to find multiple strategy profiles that support these NE i.e I see A can be supported by (A,B,C) and (A,A,A) (if there are other NE).

This is Example 1.5 in Fudenburg and Tirole.

Edit: Ok I think (B,B,B) and (C,C,C) and (A,C,C) are the other NE. I can't find anymore. Is that correct?

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    $\begingroup$ 1. The preferences here most definitely do not violate transitivity, each person has a valid utility representation. 2. You can draw the payoff matrix, but instead of a single 3x3 matrix, you will have 3 different 3x3 matrices: try it out (google three player payoff matrix if you need help). $\endgroup$ May 17 at 2:47
  • $\begingroup$ Thanks for your comment. I should have specified that collective preferences violated transitivity, which I figured had something to do because it is a majority voting game. I googled the 3x3 matrix and it seems promising. $\endgroup$
    – TirolesSon
    May 17 at 3:04

2 Answers 2

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NOTE: I assume that ties are solved randomly. So, if each voter casts a ballot for a different candidate, then each candidate gets elected with probability 1/3. So, each voter's expected payoff equals 1.

Bold numbers reflect best responses. NE means that all three numbers are in bold (sorry double stars---the markdown table is not working).

Notice that A is dominant strategy for player 1, if any other player plays A. Same for player 2 with B, and player 3 with C.

Player 3 plays A

| 1/2 | A                  |   B                 | C                   |
|----|---------------------|---------------------|---------------------|
| A  | **2**, **0**, **1** | **2**, **0**, **1** | **2**, **0**, 1 |
| B  | 2, 0, **1**         | 1, **2**, **0**     | 1, 1, 1         |
| C  | 2, 0, 1             | 1, **1**, 1         | 0, **1**, **2**     |

Player 3 plays B

| 1/2 | A              |   B                 | C               |
|----|-----------------|---------------------|-----------------|
| A  | **2**, 0, **1** | **1**, **2**, 0     | **1**, 1, 1     |
| B  | 1, **2**, 0     | **1**, **2**, **0** | **1**, **2**, 0 |
| C  | 1, 1, 1         | **1**, **2**, 0     | 0, 1, **2**         |

Player 3 plays C

| 1/2 | A              |   B                 | C               |
|----|-----------------|---------------------|-----------------|
| A  | **2**, 0, **1** | **1**, **1**, **1**     | **0**, **1**, **2** |
| B  | 1, 1, **1*      | **1**, **2**, **0** | **0**, 1, **2**     |
| C  | 0, **1**, **2** | 0, **1**, **2**     | **0**, **1**, **2**    |

All Pure Strategy Nash Eqbia:

  • A, A, A
  • A, B, A
  • B, B, B
  • A, B, C
  • B, B, C
  • A, C, C
  • C, C, C
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The proper way of solving for all NE is already specified in comments. But compared to writing down all strategy profiles and then checking, following may be a better way:

There are three types of strategy profiles:

  1. All voting for same candidate: $3$ such profiles
  2. All voting for different candidate: $6$ such profiles
  3. Two voting for same and third voting for different candidate: $18 (3\times3\times3-3-6)$ such profiles

Type 1:

Since no single player's deviation can change outcome and thus pay offs, each of these are NE.

Type 2 If each player is voting different then pay off, assuming random selection in tie, for each voter is 1. Now think of any player $i$, if any of the other player is voting for $i's$ favorite candidate then $i$ will have incentive to deviate. So only NE here can be each player playing only their favorite candidate. Hence, $(A,B,C)$

Type 3 Suppose $i,j$ are voting for same and $k$ for different. $k$ has no incentive to change. If $k$ is voting for a candidate which is either $i's$ or $j's$ favorite then the respective voter $i$ or $j$ will have incentive to deviate. So all voting in which two players play favorite and one play second favorite are all NE. So: $(A,B,A); (B,B,C); (A,C,C)$

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